A bullet moving with a velocity of 25 m/s penetrates into
a wooden log and comes to rest 0.05 s. If the mass of the
bull

2 thoughts on “A bullet moving with a velocity of 25 m/s penetrates into<br />a wooden log and comes to rest 0.05 s. If the mass of the<br />bull”

1. Answer:

   Given :-

   • A bullet moving with a velocity of 25 m/s penetrates into a wooden log and comes to rest of 0.05 seconds.
   • The mass of the bullet is 30 g.

   To Find :-

   • What is the retardation force exerted by the log on the bullet.
   • What is the depth to which the bullet penetrates.

   Formula Used :-

   [tex]\clubsuit[/tex] First Equation Of Motion :

   [tex]\longmapsto \sf\boxed{\bold{\pink{v =\: u + at}}}\\[/tex]

   where,

   • v = Final Velocity
   • u = Initial Velocity
   • a = Acceleration
   • t = Time

   [tex]\clubsuit[/tex] Force Formula :

   [tex]\longmapsto \sf\boxed{\bold{\pink{F =\: ma}}}[/tex]

   where,

   • F = Force
   • m = Mass
   • a = Acceleration

   [tex]\clubsuit[/tex] Third Equation Of Motion :

   [tex]\longmapsto \sf\boxed{\bold{\pink{v^2 – u^2 =\: 2as}}}\\[/tex]

   where,

   • v = Final Velocity
   • u = Initial Velocity
   • a = Acceleration
   • s = Distance Travelled

   Solution :-

   First, we have to find the retardation force exerted by the log on the bullet :

   Given :

   • Final Velocity = 0 m/s
   • Initial Velocity = 25 m/s
   • Time = 0.05 seconds

   According to the question by using the formula we get,

   [tex]\implies \sf 0 =\: 25 + a \times 0.05[/tex]

   [tex]\implies \sf 0 =\: 25 + 0.05a[/tex]

   [tex]\implies \sf 0 – 25 =\: 0.05a[/tex]

   [tex]\implies \sf – 25 =\: 0.05a[/tex]

   [tex]\implies \sf \dfrac{- 25}{0.05} =\: a[/tex]

   [tex]\implies \sf – 500 =\: a[/tex]

   [tex]\implies \sf \bold{\red{a =\: – 500\: m/s^2}}[/tex]

   [tex]\therefore[/tex] The retardation exerted by the log on the bullet is – 500 m/s².

   ____________________________________

   Now, we have to find the force :

   Let’s find the mass first :

   [tex]\leadsto \sf Mass = 30\: g[/tex]

   [tex]\leadsto \sf Mass =\: \dfrac{30}{1000}\: kg \: \bigg\lgroup \sf\bold{1\: g =\: \dfrac{1}{1000}\: kg}\bigg\rgroup\\[/tex]

   [tex]\leadsto \sf\bold{Mass =\: 0.03\: kg}[/tex]

   Given :

   • Mass = 0.03 kg
   • Acceleration = – 500 m/s²

   According to the question by using the formula we get,

   [tex]\implies \sf Force =\: 0.03 \times (- 500)[/tex]

   [tex]\implies \sf\bold{\green{Force =\: – 15\: N}}[/tex]

   [tex]\therefore[/tex] The retardation force exerted by the log on the bullet is 15 N .

   ___________________________

   Now, we have to find the depth to which the bullet penetrates :

   Given :

   • Final Velocity = 0 m/s
   • Initial Velocity = 25 m/s
   • Acceleration = – 500 m/s²

   According to the question by using the formula we get,

   [tex]\longrightarrow \sf (0)^2 – (25)^2 =\: 2 \times (- 500) \times s[/tex]

   [tex]\longrightarrow \sf 0 – 625 =\: – 1000 \times s[/tex]

   [tex]\longrightarrow \sf {\cancel{-}} 625 =\: {\cancel{-}} 1000s[/tex]

   [tex]\longrightarrow \sf 625 =\: 1000s[/tex]

   [tex]\longrightarrow \sf \dfrac{625}{1000} =\: s[/tex]

   [tex]\longrightarrow \sf 0.625 =\: s[/tex]

   [tex]\longrightarrow \sf\bold{\red{s =\: 0.625\: m}}[/tex]

   [tex]\therefore[/tex] The depth to which the bullet penetrates is 0.625 m .

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2. Answer :-

   → Retarding force exerted is 15 N .

   → Required depth is 0.625 m .

   Explanation :-

   We have :-

   → Initial velocity of the bullet = 25 m/s

   → Final velocity of the bullet = 0 m/s

   → Mass of the bullet = 30g = 0.03 kg

   → Time taken = 0.05 s

   ________________________________

   Firstly, let’s calculate acceleration of the bullet by using the 1st equation of motion :-

   v = u + at

   ⇒ 0 = 25 + a(0.05)

   ⇒ -25 = 0.05a

   ⇒ a = -25/0.05

   ⇒ a = -500 m/s²

   According to Newton’s 2nd law of motion :-

   F = ma

   ⇒ F = 0.03 × (-500)

   ⇒ F = -15 N

   [Here, -ve sign represents retarding force]

   ________________________________

   Now, from the 3rd equation of motion :-

   v² – u² = 2as

   ⇒ 0 – (25)² = 2(-500)s

   ⇒ -625 = -1000s

   ⇒ s = -625/-1000

   ⇒ s = 0.625 m

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